Exercises
1: Estimate fraction of molecular volume to actual volume occupied by O₂ at STP. Diameter of O₂ molecule = 3 Å.
Answer
Given: d = 3 Å = 3 × 10⁻¹⁰ m, r = 1.5 × 10⁻¹⁰ m
At STP, 1 mole of O₂ occupies Vactual = 22.4 L = 22.4 × 10⁻³ m³
Number of molecules = NA = 6.023 × 10²³
Volume of one molecule = (4/3)πr³ = (4/3) × 3.14 × (1.5 × 10⁻¹⁰)³
= (4/3) × 3.14 × 3.375 × 10⁻³⁰ = 1.414 × 10⁻²⁹ m³
Total molecular volume = NA × (volume/molecule) = 6.023 × 10²³ × 1.414 × 10⁻²⁹ = 8.52 × 10⁻⁶ m³
Fraction = 8.52 × 10⁻⁶ / (22.4 × 10⁻³) = 3.8 × 10⁻⁴
This tiny fraction confirms that molecules are mostly empty space in the gas phase.
2: Show that molar volume of ideal gas at STP is 22.4 litres.
Answer
Given: STP conditions: T = 273.15 K, P = 1 atm = 1.013 × 10⁵ Pa, R = 8.314 J mol⁻¹ K⁻¹
From ideal gas equation: PV = nRT
For n = 1 mol: V = RT/P
V = (8.314 × 273.15) / (1.013 × 10⁵)
= 2270.8 / (1.013 × 10⁵)
= 2.241 × 10⁻² m³
V = 22.41 litres ≈ 22.4 L ✓
3: PV/T vs P plots for 1.00 × 10⁻³ kg of O₂ at two temperatures T₁ and T₂.
Answer
(a) The dotted plot (horizontal line) signifies ideal gas behaviour — for a perfect ideal gas, PV/T = nR = constant regardless of pressure.
(b) At higher temperature, a real gas behaves more like an ideal gas (interactions weaker). The curve closer to the ideal dotted line corresponds to higher temperature. From the figure, the curve at T₁ is closer to ideal
⇒ T₁ > T₂.
(c) n = mass/molar mass = (1.00 × 10⁻³ kg) / (32 × 10⁻³ kg/mol) = 1/32 mol
PV/T = nR = (1/32) × 8.31 = 0.26 J K⁻¹
(d) For hydrogen (M = 2.02 g/mol): n = (1.00 × 10⁻³) / (2.02 × 10⁻³) = 0.495 mol
PV/T = 0.495 × 8.31 = 4.11 J K⁻¹ ≠ 0.26 J K⁻¹ (different value)
For same PV/T = 0.26 J K⁻¹: n = 0.26/8.31 = 0.03128 mol of H₂
Mass = 0.03128 × 2.02 = 6.32 × 10⁻² g ≈ 0.063 g of H₂
4: O₂ cylinder: volume 30 L, initial gauge pressure 15 atm at 27 °C. After withdrawal: gauge pressure 11 atm at 17 °C. Find mass of O₂ taken out. (R = 8.31 J mol⁻¹ K⁻¹, M = 32 u)
Answer
Note: Absolute pressure = gauge pressure + 1 atm
Initial: P₁ = (15 + 1) = 16 atm, T₁ = 300 K, V = 30 L = 30 × 10⁻³ m³
Final: P₂ = (11 + 1) = 12 atm, T₂ = 290 K, V = 30 L
1 atm = 1.013 × 10⁵ Pa
Initial moles: n₁ = P₁V/(RT₁) = (16 × 1.013 × 10⁵ × 30 × 10⁻³) / (8.31 × 300)
= (1.6208 × 10⁵ × 0.03) / 2493 = 4862.4 / 2493 = 1.950 mol
Final moles: n₂ = P₂V/(RT₂) = (12 × 1.013 × 10⁵ × 30 × 10⁻³) / (8.31 × 290)
= (1.2156 × 10⁵ × 0.03) / 2409.9 = 3646.8 / 2409.9 = 1.513 mol
Moles withdrawn = 1.950 − 1.513 = 0.437 mol
Mass = 0.437 × 32 = 13.98 g ≈ 141 g
More precisely: Δn × 32 = 0.437 × 32 ≈ 13.98 g
5: Air bubble volume 1.0 cm³ at bottom of 40 m deep lake at 12 °C. Volume at surface (35 °C)?
Answer
At bottom: P₁ = P₀ + ρgh = 1.013 × 10⁵ + 1000 × 9.8 × 40 = 1.013 × 10⁵ + 3.92 × 10⁵ = 4.933 × 10⁵ Pa
T₁ = 12 + 273 = 285 K, V₁ = 1.0 cm³
At surface: P₂ = 1.013 × 10⁵ Pa, T₂ = 35 + 273 = 308 K
Using combined gas law: P₁V₁/T₁ = P₂V₂/T₂
V₂ = V₁ × (P₁/P₂) × (T₂/T₁) = 1.0 × (4.933/1.013) × (308/285)
= 1.0 × 4.870 × 1.0807 = 5.26 cm³
6: Estimate total number of air molecules in a room of 25.0 m³ at 27 °C and 1 atm pressure.
Answer
Given: V = 25.0 m³, T = 300 K, P = 1.013 × 10⁵ Pa
Number of moles: n = PV/(RT) = (1.013 × 10⁵ × 25.0) / (8.314 × 300)
= 2.5325 × 10⁶ / 2494.2 = 1015.4 mol
N = n × NA = 1015.4 × 6.023 × 10²³ = 6.117 × 10²⁶ ≈ 6.1 × 10²⁶ molecules
7: Average thermal energy of a helium atom at: (i) 27 °C, (ii) 6000 K (Sun surface), (iii) 10⁷ K (star core).
Answer
Average kinetic energy = (3/2)kBT, where kB = 1.38 × 10⁻²³ J/K
(i) T = 300 K: E = (3/2) × 1.38 × 10⁻²³ × 300 = 6.21 × 10⁻²¹ J
(ii) T = 6000 K: E = (3/2) × 1.38 × 10⁻²³ × 6000 = 1.241 × 10⁻¹⁹ J
(iii) T = 10⁷ K: E = (3/2) × 1.38 × 10⁻²³ × 10⁷ = 2.07 × 10⁻¹⁶ J
8: Three vessels (equal capacity, same T, P) with Ne (monatomic), Cl₂ (diatomic), UF₆ (polyatomic). Equal molecules? Same vrms? Largest vrms?
Answer
Equal number of molecules? Yes. At same T and P with equal volume, PV = NkT
⇒ N = PV/(kT) is same for all three. Each vessel contains the same number of molecules (Avogadro's hypothesis).
Same vrms? No. vrms = √(3RT/M) depends on molar mass M.
Molar masses: Ne = 20 g/mol, Cl₂ = 71 g/mol, UF₆ = 352 g/mol
Smallest M
⇒ largest vrms. Neon has the largest vrms.
9: At what temperature is vrms of Ar equal to vrms of He at −20 °C? MAr = 39.9 u, MHe = 4.0 u.
Answer
vrms = √(3RT/M)
⇒ For equal vrms: TAr/MAr = THe/MHe
TAr = THe × (MAr/MHe) = 253 × (39.9/4.0) = 253 × 9.975 = 2523.4 K ≈ 2523 K
10: Estimate mean free path and collision frequency of N₂ at 2.0 atm and 17 °C. Radius of N₂ molecule ≈ 1.0 Å, M = 28.0 u.
Answer
Given: P = 2 × 1.013 × 10⁵ = 2.026 × 10⁵ Pa, T = 290 K, r = 1.0 × 10⁻¹⁰ m, d = 2r = 2 × 10⁻¹⁰ m
Number density: n = P/(kBT) = 2.026 × 10⁵ / (1.38 × 10⁻²³ × 290) = 2.026 × 10⁵ / 4.002 × 10⁻²¹ = 5.06 × 10²⁵ m⁻³
Mean free path: λ = 1 / (√2 × π × d² × n)
= 1 / (1.414 × 3.14 × (2 × 10⁻¹⁰)² × 5.06 × 10²⁵)
= 1 / (1.414 × 3.14 × 4 × 10⁻²⁰ × 5.06 × 10²⁵)
= 1 / (1.414 × 3.14 × 2.024 × 10⁶)
= 1 / (8.984 × 10⁶) = 1.11 × 10⁻⁷ m ≈ 1.1 × 10⁻⁷ m
RMS speed: vrms = √(3RT/M) = √(3 × 8.31 × 290 / 0.028) = √(258,246) = 508 m/s
Mean speed vavg ≈ 0.921 × vrms ≈ 468 m/s
Collision frequency: f = vavg/λ = 468 / (1.11 × 10⁻⁷) = 4.2 × 10⁹ s⁻¹
Collision time: tcollision ≈ d/vavg = 2 × 10⁻¹⁰ / 468 = 4.3 × 10⁻¹³ s
Free travel time: tfree = λ/vavg = 1.11 × 10⁻⁷ / 468 = 2.4 × 10⁻¹⁰ s
Ratio tfree/tcollision ≈ 2.4 × 10⁻¹⁰ / 4.3 × 10⁻¹³ ≈ 558 ≈ 500
The molecule spends about 500 times longer in free travel than in collision — confirming that collisions are brief events.
Important Extra Questions
Extra Q1: State the assumptions of the kinetic theory of gases. Which assumptions break down for real gases?
Answer:
Assumptions of kinetic theory
- A gas consists of a large number of molecules in random motion.
- Molecules are point masses — their volume is negligible compared to the volume of the container.
- Intermolecular forces are negligible except during collisions.
- Collisions are perfectly elastic — kinetic energy is conserved.
- All directions of motion are equally probable (isotropy).
- The time of a collision is negligible compared to the time between collisions.
Assumptions that break down for real gases: Assumptions 2 and 3 — real molecules have finite volume and exert attractive/repulsive forces at short distances. These are accounted for in the van der Waals equation: (P + a/V²)(V − b) = nRT.
Extra Q2: Derive the expression for pressure exerted by an ideal gas using kinetic theory.
Answer:
Consider N molecules in a cubical box of side L. For a molecule with x-velocity vx:
Momentum change per collision with wall = 2mvx
Time between successive collisions with same wall = 2L/vx
Force by one molecule = 2mvx / (2L/vx) = mvx²/L
Total force by N molecules: F = (Nm/L)<vx²>
Since motion is isotropic: <vx²> = <v²>/3
Pressure P = F/L² = Nm<v²>/(3L³) = Nm<v²>/(3V)
P = (1/3)(N/V)m<v²> = (1/3)ρ<v²>
Comparing with PV = NkT: <(1/2)mv²> = (3/2)kBT — the average kinetic energy per molecule.
Extra Q3: What is the law of equipartition of energy? How does it predict the specific heat ratios?
Answer:
Law of equipartition of energy: In thermal equilibrium, each degree of freedom of a molecule contributes (1/2)kBT to its average energy.
Degrees of freedom (f): Monatomic = 3; Diatomic (rigid) = 5 (3 trans + 2 rot); Diatomic (non-rigid) = 7 (+ 2 vib); Polyatomic = 6 or more.
CV = (f/2)R, CP = CV + R, γ = CP/CV = 1 + 2/f
Monatomic: γ = 1 + 2/3 = 5/3 ≈ 1.67 (agrees with He, Ar)
Diatomic (rigid): γ = 1 + 2/5 = 7/5 = 1.4 (agrees with H₂, O₂, N₂)
Extra Q4: What is Brownian motion? What does it demonstrate about the nature of heat?
Answer:
Brownian motion is the random, irregular zigzag motion of small particles (e.g., pollen grains, smoke particles) suspended in a fluid (liquid or gas), first observed by botanist Robert Brown in 1827.
Cause: The particles are constantly bombarded by the fast-moving molecules of the surrounding fluid. Due to random fluctuations, the impulses from different directions don't exactly cancel, causing the particle to move erratically.
Significance
- Direct evidence for the existence and motion of molecules.
- Confirms that temperature is a measure of average molecular kinetic energy.
- Einstein (1905) provided a quantitative theory of Brownian motion, enabling the first accurate determination of Avogadro's number.
Extra Q5: The rms speed of hydrogen molecules at 27 °C is approximately 1934 m/s. Compare this with Earth's escape velocity (11.2 km/s). Can hydrogen escape Earth's atmosphere?
Answer:
vrms(H₂) = √(3RT/M) = √(3 × 8.31 × 300 / 0.002) = √(3,739,500) = 1934 m/s
Escape velocity = 11,200 m/s
vrms / vescape = 1934/11200 ≈ 0.17
Although the average speed is much less than escape velocity, the Maxwell speed distribution has a high-speed tail. A fraction of molecules have speeds exceeding the escape velocity — especially in the upper atmosphere where temperature can be higher.
Over geological time scales, these fast molecules escape, which is why Earth has very little hydrogen in its atmosphere today. Heavier gases (N₂, O₂) have much lower vrms (~500 m/s) and virtually no molecules reach escape velocity — so they are retained.